JEE PYQ: Trigonometric Equations And Inequations Question 12
Question 12 - 2019 (10 Apr Shift 1)
All the pairs $(x, y)$ that satisfy the inequality $2\sqrt{\sin^2 x - 2\sin x + 5} \cdot \frac{1}{4\sin^2 y} \le 1$ also satisfy the equation:
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(1) $2|\sin x| = 3 \sin y$
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(2) $2 \sin x = \sin y$
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(3) $\sin x = 2 \sin y$
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(4) $\sin x = |\sin y|$
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Answer:
Solution
$\sqrt{(\sin x - 1)^2 + 4} \le 2\sin^2 y$
True if $\sin x = 1$ and $|\sin y| = 1$, therefore $\sin x = |\sin y|$.
BETA
